Hi!
(x+a)·(x+b) = x²+xa+xb+ab = x²-12x-45
x² = x²
xa+xb = (a+b)x = -12x ⇒ (a+b) = -12
ab = -45
When (a+b) = -12 and ab = -45?
+3-15 = -12 and (+3)(-15) = -45
-3-15 = -18 and (-3)(-15) = +45
+3+15 = +18 and (+3)(+15) = +45
-3+15 = +12 and (-3+15) = -45
Answer:
(x+3)(x-15)
<span>Point B has coordinates (3,-4) and lies on the circle. Draw the perpendiculars from point B to the x-axis and y-axis. Denote the points of intersection with x-axis A and with y-axis C. Consider the right triangle ABO (O is the origin), by tha conditions data: AB=4 and AO=3, then by Pythagorean theorem:
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</span>
{Note, that BO is a radius of circle and it wasn't necessarily to use Pythagorean theorem to find BO}
<span>The sine of the angle BOA is</span>

Since point B is placed in the IV quadrant, the sine of the angle that is <span> drawn in a standard position with its terminal ray will be </span>
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</span><span>
</span>

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Answer:
1
Step-by-step explanation:
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